(Timothy Bodenmiller) wrote:
TB> As far as the water canopy goes, I will have to post a theory on
TB> how a liquid canopy could stay above the earth. There are several
TB> things that must be accounted for in order to see if a sphere could
TB> be formed. The sphere of liquid water must be placed at exactly the
TB> right distance, for the gravitational pull to be right. Also,
TB> water's molecular attraction must be taken into account. I will do
TB> some calculations on the approximate volume range that the water
TB> canopy would have to have had, and then go from there. The canopy
TB> does not necessarily need to be placed so far away from the earth
TB> that it can only exist as ice, also, depending on the width of the
TB> canopy, and assuming that it was pure water, light should get
TB> through o.k.
From: email@example.com (glenn r morton) wrote back:
You want a liquid canopy in orbit. Such a situation will not
accomplish what you are desiring- namely a source for the flood
waters. Here is why. If you place liquid water in space, where there
is no pressure, it will immediately begin to boil (check any steam
table) If I am not mistaken, water boils at approximately 140
degrees F in Denver so a three minute egg there is quite raw. I
will rehash the orbital problems that other responders have already
mentioned. A sphere is an impossible object to orbit around a
planet. Any two objects in mutual orbit around each other (water
molecule and earth in this case) must revolve around the center of
mass. Szebehely, _Adventures in Celestial Mechanics_, Austin:
University of Texas Press, 1989, p. 4, states;
"The satellite and the Earth are moving around the center of mass of
the Earth-satellite system. Since the mass of the Earth is always
many orders of magnitude larger than the mass of the satellite, the
center of mass of the system is at the center of the Earth for all
Buckle your seat belt, boy, cause you are going to need it. Let's
consider two 1 gram drops of water in orbit: one in polar orbit and
one in equatorial orbit. If the orbits are at the same height, and
the timing is correct, eventually, the two drops will collide. What
happens in this collision. In order to see, you need to ride on one
of the drops. To you, your drop is stationary it is the other one
that is moving. Due to the geometry of the situation (see below).
For orbital velocity calculation see Szebehely, p. 2. The drop
coming at you has two components of velocity - 5 mi/sec directly
towards you and 5 mi/sec transverse to you.
5 mi /sec
By the Pythagorean theorem the direct velocity is sqrt(5*5+5*5)=
approximately 7 miles per sec. That has got to be a scary site
seeing a 1 gram water droplet coming at you with that speed!
Converting this the meters per second yields 11268 meters / sec.
Have you ever been hit with a water balloon thrown from a car going
40 miles an hour? I have--- it hurts like hell and yet that balloon
was only traveling 18 meters per second. Have you ever heard of
kinetic energy, Mr. Bodenmiller? The equation is KE=.5*m*v*v. The
drop that is coming towards you has 63483 joules of energy or 15172
calories. The two one-gram water droplets will be instantly vaporized
and the resultant vapor would be around 60,000 degrees K. Water
vapor has an approximate heat capacity of .25 calories per degree C.
Remember the heat of vaporization is 597 calories. Thus a stream of
particles in polar orbit will be traveling at right angles to the
stream of water molecules in equatorial orbit. If they are at the
same height, they will collide and vaporize due to the energy of the
collision. The other posters are quite correct about this and your
mechanism for a sphere is wrong for precisely this reason.
Let us assume that you are correct ("everybody else is wrong") and
you can place a sphere of water in orbit. What happens when it falls
to the earth as rain? Ignoring the kinetic energy of the orbital
motion there is still a problem. Assume God stops the orbiting
liquid and just lets it fall. There is a famous equation in physics
describing the potential energy of an object. E=mgh, where E is the
energy, m is the mass g is the acceleration of gravity and h is the
height. Although there are minor variations in g as one ascends to
orbital heights, for simplicity we will consider g constant. Take 1
gram of water at 240 km (150 miles). The energy released when the
water falls to the earth is 1 gm * 24,000,000 cm * 980 cm/sec/sec
=23,520,000,000 ergs = 562 calories. What does this mean? Well it is
enough energy to raise the temperature of water from 0 to 100 deg. C
and then boil off over 2/3 of the water. The heat of vaporization
for water is approximately 597 calories per gram. (see John T.
Houghton _Physics of Atmospheres_ Cambridge, 1977), appendix 2 p.
165) I will leave it to you to convert from joules to calories
(although I am not sure that I really ought to). Thus since it takes
1 calorie per degree, if your water started at 0 deg. C when it hits
the earth the temperature of it is now 100 deg C and there is 462
calories left. These 462 calories will go into boiling off the
water. The precise fraction is 462/597=.77 will become steam.
How well do you think the plants floating in the boiling flood
waters will fare? How about Noah and the animals on the ark. They
are bathed in hot scalding steam. This is a means of cooking food
you know--- steaming.
Even ignoring this problem you have a major problem with radiative
heat transfer. When sunlight strikes the earth, part of the energy
is absorbed and part is reflected back into space. The fraction
which is reflected back to space is called the albedo. The amount of
energy absorbed from the sun each minute is
where Ea is the energy absorbed, pi=3.14, R is the earth's radius, S
is the solar constant (2 calories/minute/cm/cm) and A is the albedo
(.36 for the earth). See equation 4-1 R. M. Goody and J.C. G.
Walker, _Atmospheres_, Prentice-Hall, 1972, p. 88. The pi R squared
term is due to the fact that only half of the earth is sunlit and it
is sunlit from only one direction. However when it comes to emission
of energy, each square centimeter of the earth's surface emits
energy at all times. Thus
where Ee is the emitted energy, and F is the infrared flux leaving
the earth. Remember that the absorbed sunlight is re-emitted in the
infrared region primarily between 10 and 13 microns wavelength.
Water vapor has very little absorption in that bandwidth so the
energy escapes to space and we remain comfortable. Unfortunately
that is NOT true with liquid water which you have in your canopy!!!!
Liquid water absorbs strongly in the 10-13 micron band. If you put
liquid water surrounding the earth the heat will build up very
quickly to Venusian levels.
For radiative equilibrium Ee=Ea. If this is not true then either the
earth is cooling off or heating up. Solving for the flux, F, after
substituting the respective equations for Ee and Ea we have
If I were to explain the next equation I would put most people to
sleep. The radiative heat transfer equation for an atmosphere is
where tau is the optical depth, B(tau) is the blackbody function,
sigma*T^4 , and sigma is the Stephan-Boltzman constant. (See V. A.
Ambartsumyan, Theoretical Astrophysics, New York Pergamon Press,
1958 , p. 23-25 for details) Solving for T^4 and substituting the
flux F equation into the B(tau) equation gives
T^4 =(S(1-A)(1+1.5*tau))/(8*sigma) final equation!!!!
With the fact that 1 cm of precipitable water in a water vapor
canopy is approximately equal to an optical depth of 1, if you place
even [as little as] 30 cm of precipitable water vapor surrounding
the earth, the temperature at the present albedo is around 543
degrees K or 270 degrees C or 518 degrees F. It is so hot that Adam
and Eve would have had to go around naked. Seriously, all life would
die. All you could get out of this canopy is a flood one foot (30
cm) in depth! This is hardly enough water to deposit 60,000 feet of
sediment in the Anadarko Basin of Oklahoma!
Your liquid canopy even makes the problem worse. Water vapor is a
strong infrared absorber at nearly all frequencies except 10-12
microns. Our existence is due to the coincidence that the earth's
infrared emissions peak in this window. However, liquid water
absorbs much more strongly in the 10-12 micron window making it
harder for the heat to escape. This can be observed on cold winter
nights. Generally speaking the coldest nights are cloudless and dry.
Night time lows do not get as low when a blanket of clouds helps
retain the heat at the earth's surface.
One final piece of physics. Are you aware that the average lifetime
for a water molecule exposed to the ultraviolet light of the sun is
somewhere between 20 and 81 hours? Take a look at A. E. Potter and
Betty Del Duca, "Lifetime in Space of Possible Parent Molecules of
Cometary Radicals", Icarus (3), 1964, p. 105. Your space water vapor
canopy would be turned into hydrogen and oxygen in a short space of
A word of advice to you Chris, if you do not understand any of this
physics, then you better study it before you try to debate it.
If you have not had orbital mechanics, what are you doing proposing
If you have not had frosh physics, what are you doing suggesting
rainfall from 240 km high? Is it because someone else suggested it?
Did they do the math?
If you have not had thermodynamics what are you doing proposing to
place water in orbit with no thought for the thermodynamic behavior
of the material?
If you have not had radiative heat transfer what are you doing
suggesting a canopy?
Deal with the physics by means of mathematics not verbiage.
Mathematics is the language of science not English. But doing
science with English seems to be the creationist style of physics!
I did not see any response to my note about the spherical shape of
the earth and its implication for your claim that rain did not fall
before the flood. Can you not even deal with geometry?
You are being duped by those whom you trust, namely the leaders of
the Creationist movement. Have you ever heard of Godel's theorm?
Basically it says that any system of thought devised by man will
have issues which can not be resolved within that system. They will
always remain problematical and unanswerable. Ask your professors if
there is any problems that creationism has that don't seem to be
able to be answered. If they tell you that there are none, then you
will know that they either don't understand what they are teaching
or they are deceptive! (Evolutionists should note that they also
suffer occassionally from this disease.) I'll bet you won't do this.
Better still go ask them how there could be worm burrows throughout
the sediments of the Anadarko Basin when the sedimentation rate
during the flood was 60,000/365= 164 feet per day. That is nearly 7
feet per hour. How rapidly did those worms make their burrows. I bet
your prof will say that there really aren't any worm burrows.
Now as for you Jim Lippard. Before you drag me into this again at
least have the courtesy to send me your address, like I asked you
to, so I can converse with you outside of electronic means.
From: Robert J. Kolker
To: All Msg #102, Jan-31-94 06:15AM
Subject: Re: Vapor Canopy
Organization: The World Public Access UNIX, Brookline, MA
From: firstname.lastname@example.org (Robert J. Kolker)
I loved your refutation of the Flood, using mathematical analysis
and physical theory. I have only one nit to pick with your otherwise
flawless presentation. You got Goedel's theorem wrong. Goedel's
theorm says that if first order logic + arithmetic axioms are
consistent there exists a formula F such that neither F nor -F is
provable. There are many mathematical systems that are completely
decidable. For example, propositional logic is completely decidable.
Just use truth tables. Any Well Formed Formual which is tautological
(all T values in the table) is provable and conversely.
BTW, the Creationists will tell you that the laws of physics have
changed since the time of the flood so your arguments do not carry.
Keep up the good work.
Conan the Libertarian